Question

A typical electrical generating plant has a capacity of $\mathbf{500}\mathbf{MW}$($\mathbf{1}\mathbf{}\mathbf{MW}\mathbf{=}\mathbf{1}{\mathbf{Js}}^{\mathbf{-}\mathbf{1}}$) and an overall efficiency of about $\mathbf{25}\mathbf{\%}$.

(a) The combustion of 1 kg of bituminous coal releases about $\mathbf{3}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{KJ}$and leaves an ash residue of $\mathbf{100}\mathbf{}\mathbf{g}$. What weight of coal must be used to operate a role="math" localid="1661450615107" $\mathbf{500}\mathbf{MW}$generating plant for 1 year, and what weight of ash must be disposed of?

(b) Enriched fuel for nuclear reactors contains about $\mathbf{4}\mathbf{\%}$, fission of which gives $\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{KJ}\mathbf{}\mathbf{per}\mathbf{}\mathbf{mole}$. What weight of is needed to operate a $\mathbf{500}\mathbf{MW}$power plant, assumed to have $\mathbf{25}\mathbf{\%}$efficiency, for 1 year, and what weight of fuel must be reprocessed to remove radioactive wastes?

(c) The radiation from the sun striking the earth’s surface on a sunny day corresponds to a power of . How large must the collection surface be for a $\mathbf{500}\mathbf{MW}$solar generating plant? (Assume that there are role="math" localid="1661450903101" $\mathbf{6}$hours of bright sun each day and that storage facilities continue to produce power at other times. The efficiency for solar-power generation would be about $\mathbf{25}\mathbf{\%}$.)

Short answer

1. The weight of coal used is $1.875\times {10}^{11}\mathrm{kg}$and the weight of ash disposed is $1.875\times {10}^{13}\mathrm{g}$.
2. The weight of uranium used is data-custom-editor="chemistry" $7.46\times {10}^7\mathrm{g}$. the weight of fuel that needs to be reprocessed data-custom-editor="chemistry" $1.865\times {10}^9\mathrm{g}$.
3. The area of the solar panel is$4.0\times {10}^4{\mathrm{m}}^2$

Step-by-step solution

Capacity of the power plant

Capacity of the power plant $=500\mathrm{MW}=5\times {10}^8{\mathrm{Js}}^{-1}$

Energy consumed in a yearrole="math" localid="1661451374951" $=5\times {10}^8\times 365\times 24\times 3600\mathrm{J}=1.5\times {10}^{18}\mathrm{J}$

Weight of coal used

Let the weight of coal be $\mathrm{x}\mathrm{kg}$.

Energy released by $\mathrm{x}\mathrm{kg}$of coal $=\mathrm{x}\times 3.2\times {10}^4\mathrm{KJ}=3.2\mathrm{x}\times {10}^7\mathrm{J}$

Since the powerplant is 25% efficient.

$$\begin{gathered} \frac{25}{100}\times \mathrm{x}\times 3.2\times {10}^{7}J=1.5\times {10}^{18}\mathrm{J} \\ \Rightarrow \mathrm{x}=\frac{1.5}{3.2}\times 4\times {10}^{11}\mathrm{kg} \\ \Rightarrow \mathrm{x}=1.875\times {10}^{11}\mathrm{kg} \end{gathered}$$

Weight of coal consumed $=1.875\times {10}^{11}\mathrm{kg}$

1 kg of coal produce 100 gm ash residue

Therefore,

Amount of ash residue produced $=\frac{1.875\times {10}^{11}\mathrm{kg}}{1\mathrm{kg}}\times 100\mathrm{g}=1.875\times {10}^{13}\mathrm{g}$

Weight of uranium used

Amount of energy released$1.9\times {10}^{10}\mathrm{kJ}=1.9\times {10}^{13}\mathrm{J}$

Let the number of moles of uranium needed be y.

Since the engine is only 25% efficient,

localid="1661507913832" $\begin{array}{c}\frac{25}{100}\times \mathrm{y}\times 1.9\times {10}^{13}=1.5\times {10}^{18}\\ \mathrm{y}=\frac{1.5\times 4}{1.9}\times {10}^5\mathrm{mol}\\ \mathrm{y}=3.175\times {10}^5\mathrm{mol}\end{array}$

Mass of uranium consumed localid="1661507988706" width="256">=3.175×105×235g=7.46×107g

Since, the enriched fuel consists of only 4% uranium,

Amount of fuel that needs to be reprocessed$=7.46x{10}^{7}x\frac{100}{4}g=1.865x{10}^9\mathrm{g}$

Solar energy

The efficiency of the solar power plant is 25%.

Therefore,

Amount of energy needed to be produced to run a 500 MW power station per day$5\times {10}^6\times 18\times 3600=3.25\times {10}^{11}\mathrm{J}$

Since the solar power plant is 25% efficient,

Total energy produced by the plant$=3.25\times {10}^{11}x\frac{100}{25}=13x{10}^{11}\mathrm{J}$

Area of the collection surface$=\frac{13\times {10}^{11}\mathrm{J}}{1.5\times {10}^3\times 6\times 3600{\mathrm{Jm}}^{-2}}=4\times {10}^4{\mathrm{m}}^2$